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CBSE Class 11-science Answered

What will be the answer of question 27 ?

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Asked by arunavamitra50 | 16 Jun, 2018, 06:08: PM

Expert Answer

Let the stone has taken t seconds to reach the ground.

At (t-1)^th second , speed of stone = g(t-1) m/s

Hence distance travelled at last 1 second is given by,

begin mathsize 12px style fraction numerator 9 h over denominator 25 end fraction space equals space g left parenthesis t minus 1 right parenthesis space plus 1 half g space equals space g left parenthesis t minus 1 half right parenthesis space....................... left parenthesis 1 right parenthesis end style

(To get eqn.(1), we used the formula " S = ut+(1/2)gt²", and t = 1 is substituted since it is only one second travel )

for complete travel of height h, we have,

begin mathsize 12px style h space equals space 1 half g space t squared space space........................... left parenthesis 2 right parenthesis end style

let us substitute h from eqn.(2) in eqn.(1), and after simplification, eqn.(1) is modified as given below

begin mathsize 12px style fraction numerator 9 t squared over denominator 50 end fraction minus space t plus 1 half space equals space 0 space space o r space 9 t squared minus 50 t plus 25 space equals space 0 end style

.................................(3)

solving for t using eqn.(3), we get t = 5 seconds.

Hence complete travel of height h will be obtained from eqn.(2) and it is given by

h = (1/2)×10×5×5 = 125 m

Answered by Thiyagarajan K | 16 Jun, 2018, 11:10: PM

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